Question

1. a circle in the xy - plane has equation $(x - 4)^{2}+(y + 3)^{2}=36$. which of the following points does not lie in the interior of the circle?

(a) $(-1,-5)$
(b) $(-2,0)$
(c) $(0,0)$
(d) $(8,-5)$

Explanation:

Step1: Recall the equation of a circle

The standard form of a circle's equation is \((x - h)^2+(y - k)^2=r^2\), where \((h,k)\) is the center and \(r\) is the radius. For the circle \((x - 4)^2+(y + 3)^2 = 36\), the center is \((4,-3)\) and \(r=\sqrt{36}=6\).

Step2: Use the distance formula

The distance formula between two points \((x_1,y_1)\) and \((x_2,y_2)\) is \(d=\sqrt{(x_2 - x_1)^2+(y_2 - y_1)^2}\). If \(d\lt r\), the point \((x_2,y_2)\) lies inside the circle.

For point \((-1,-5)\)

\(d=\sqrt{(4-(-1))^2+(-3-(-5))^2}=\sqrt{(5)^2+(2)^2}=\sqrt{25 + 4}=\sqrt{29}\approx5.39\lt6\)

For point \((-2,0)\)

\(d=\sqrt{(4-(-2))^2+(-3 - 0)^2}=\sqrt{(6)^2+(-3)^2}=\sqrt{36+9}=\sqrt{45}\approx6.71\gt6\)

For point \((0,0)\)

\(d=\sqrt{(4 - 0)^2+(-3-0)^2}=\sqrt{16 + 9}=\sqrt{25}=5\lt6\)

For point \((8,-5)\)

\(d=\sqrt{(4 - 8)^2+(-3-(-5))^2}=\sqrt{(-4)^2+(2)^2}=\sqrt{16+4}=\sqrt{20}\approx4.47\lt6\)

Answer:

B. \((-2,0)\)